= Solution
For $|z|<1$, multiply the absolutely convergent <Taylor series> of the sine by the <geometric series>:
$$
\frac{\sin z}{1-z^2}=\left[\sum_{j=0}^\infty\frac{(-1)^jz^{2j+1}}{(2j+1)!}\right]\left[\sum_{k=0}^\infty z^{2k}\right].
$$
The product is odd, so all even coefficients vanish. Collecting the coefficient of $z^{2m+1}$ gives
$$
\boxed{a_{2m}=0,\qquad a_{2m+1}=\sum_{j=0}^m\frac{(-1)^j}{(2j+1)!},\qquad m=0,1,2,\ldots.}
$$
For example, the first terms are $z+5z^3/6+101z^5/120+\cdots$. The denominator vanishes at $z=\pm1$, and the numerator is nonzero there, so these are genuine <simple poles>. Therefore $\boxed{R=1}$ is the <radius of convergence>.
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